The Unapologetic Mathematician

Distinct Eigenvalues

Today I’d like to point out a little fact that applies over any field (not just the algebraically-closed ones). Let $T$ be a linear endomorphism on a vector space $V$, and for $1\leq i\leq n$, let $v_i$ be eigenvectors with corresponding eigenvalues $\lambda_i$. Further, assume that $\lambda_i\neq\lambda_i$ for $i\neq j$. I claim that the $v_i$ are linearly independent.

Suppose the collection is linearly dependent. Then for some $k$ we have a linear relation

$\displaystyle v_k=c_1v_1+c_2v_2+...+c_{k-1}v_{k-1}$

We can assume that $k$ is the smallest index so that we get such a relation involving only smaller indices.

Hit both sides of this equation by $T$, and use the eigenvalue properties to find

$\displaystyle\lambda_kv_k=c_1\lambda_1v_1+c_2\lambda_2v_2+...+c_{k-1}\lambda_{k-1}v_{k-1}$

On the other hand, we could just multiply the first equation by $\lambda_k$ to get

$\displaystyle\lambda_kv_k=c_1\lambda_kv_1+c_2\lambda_kv_2+...+c_{k-1}\lambda_kv_{k-1}$

Subtracting, we find the equation

$\displaystyle0=c_1(\lambda_k-\lambda_1)v_1+c_2(\lambda_k-\lambda_2)v_2+...+c_{k-1}(\lambda_k-\lambda_{k-1})v_{k-1}$

But we this would contradict the minimality of $k$ we assumed before. Thus there can be no such linear relation, and eigenvectors corresponding to distinct eigenvalues are linearly independent.

February 4, 2009 - Posted by | Algebra, Linear Algebra

1. […] vectors here, they span a subspace of dimension , which must be all of . And we know, by an earlier lemma that a collection of eigenvectors corresponding to distinct eigenvalues must be linearly […]

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2. […] Eigenspaces are Disjoint We know that eigenspaces of distinct eigenvalues are disjoint. That is, no vector is an eigenvector of two different eigenvalues. This extends to generalized […]

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3. […] as a corollary we can find that not only are the eigenvectors corresponding to distinct eigenvalues linearly independent, they are actually orthogonal! Indeed, if and are eigenvectors of with […]

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4. Is the dimension of V, n? ie dim(V)=n, ie their contains n elements in a basis for V. since you have n eigenvectors that are linearly independent and any set of n linearly independent vectors is a basis for an n dimensional vectorspace.

Comment by H | April 18, 2010 | Reply

5. Yes, H, that’s true. But all this post was intended to prove was their linear independence.

Oh, and I think you mean “there”.

Comment by John Armstrong | April 19, 2010 | Reply